Make a table...

**jzellt** · March 23rd 2010, 08:48 PM

Make a table showing all the values of the functions u4 : Z15 ->Z15.

The function u

a defined by the rule x ->ax is a bijection Zm ->Zm.

How is this done? Thanks...

**chiph588@** · March 23rd 2010, 09:19 PM

Originally Posted by jzellt

Make a table showing all the values of the functions u4 : Z15 ->Z15.

The function u

a defined by the rule x ->ax is a bijection Zm ->Zm.

How is this done? Thanks...

$(4,15)=1 \implies \{4\cdot1, 4\cdot2,\cdot\cdot\cdot, 4\cdot 15\} = \{1,2,\cdot\cdot\cdot,15\}$

Therefore $u_4(\mathbb{Z}_{15}) = \mathbb{Z}_{15}$ .

$\begin{array}{r|r||r|r}a&u_4(a)&a&u_4(a)\\ \hline 0&0&8&2\\1&4&9&6\\2&8&10&10\\3&12&11&14\\4&1&12&3\ \5&5&13&7\\6&9&14&11\\7&13\end{array}$

**jzellt** · March 23rd 2010, 09:25 PM

Thanks for the quick reply, but I'm still not seeing what my table should look like...

**chiph588@** · March 23rd 2010, 09:28 PM

Originally Posted by chiph588@

$\begin{array}{r|r||r|r}a&u_4(a)&a&u_4(a)\\ \hline 0&0&8&2\\1&4&9&6\\2&8&10&10\\3&12&11&14\\4&1&12&3\ \5&5&13&7\\6&9&14&11\\7&13\end{array}$

Here you go.

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